\(\frac{3x+1}{x+y}-\frac{2x-3}{x+y}\)

\(=\frac{3x+1-2x+3}{x+y}\)

\(=\frac{x+4}{x+y}\)

giúp mik vs

\(\frac{2\left(x+y\right)\left(x-y\right)}{x}-\frac{-2y^2}{x}\)

\(=\frac{2\left(x^2-y^2\right)+2y^2}{x}\)

\(=\frac{2x^2-2y^2+2y^2}{x}\)

\(=\frac{2x^2}{x}\)

\(=2x\)

\(\frac{1-3x}{2}-\frac{x+3}{2}\)

\(=\frac{1-3x-x-3}{2}\)

\(=\frac{-4x-2}{2}\)

\(=-2x-1\)

a)  \(4\left(2x+7\right)^2=9\left(x+3\right)^2\)

\(\Leftrightarrow4\left(4x^2+28x+49\right)=9\left(x^2+6x+9\right)\)

\(\Leftrightarrow16x^2+112x+196=9x^2+54x+81\)

\(\Leftrightarrow7x^2+58x+115=0\)

\(\Leftrightarrow7x^2+35x+23x+115=0\)

\(\Leftrightarrow7x\left(x+5\right)+23\left(x+5\right)=0\)

\(\Leftrightarrow\left(x+5\right)\left(7x+23\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x+5=0\\7x+23=0\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=-5\\x=-\frac{23}{7}\end{cases}}\)

Vậy tập nghiệm của phương trình là \(S=\left\{-5;-\frac{23}{7}\right\}\)

b) \(2x^3+7x^2+7x+2=0\)

\(\Leftrightarrow2x^3+2x^2+5x^2+5x+2x+2=0\)

\(\Leftrightarrow2x^2\left(x+1\right)+5x\left(x+1\right)+2\left(x+1\right)=0\)

\(\Leftrightarrow\left(x+1\right)\left(2x^2+5x+2\right)=0\)

\(\Leftrightarrow\left(x+1\right)\left(2x^2+4x+x+2\right)=0\)

\(\Leftrightarrow\left(x+1\right)\left[2x\left(x+2\right)+\left(x+2\right)\right]=0\)

\(\Leftrightarrow\left(x+1\right)\left(2x+1\right)\left(x+2\right)=0\)

\(\Leftrightarrow\)\(x+1=0\)

hoặc \(2x+1=0\)

hoặc  \(x+2=0\)

\(\Leftrightarrow\) \(x=-1\)

hoặc    \(x=-\frac{1}{2}\)

hoặc    \(x=-2\)

 Vậy tập nghiệm của phương trình là \(S=\left\{-1;-\frac{1}{2};-2\right\}\)

c) \(x^4+x^2+6x-8=0\)

\(\Leftrightarrow x^4-x^3+x^3-x^2+2x^2-2x+8x-8=0\)

\(\Leftrightarrow x^3\left(x-1\right)+x^2\left(x-1\right)+2x\left(x-1\right)+8\left(x-1\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left(x^3+x^2+2x+8\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left(x^3+2x^2-x^2-2x+4x+8\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left[x^2\left(x+2\right)-x\left(x+2\right)+4\left(x+2\right)\right]=0\)

\(\Leftrightarrow\left(x-1\right)\left(x+2\right)\left(x^2-x+4\right)=0\)

\(\Leftrightarrow\)\(x-1=0\)

hoặc   \(x+2=0\)

hoặc   \(x^2-x+4=0\)

\(\Leftrightarrow\)\(x=1\)(tm)

hoặc   \(x=-2\)(tm)

hoặc  \(\left(x-\frac{1}{2}\right)^2+\frac{15}{4}=0\)(ktm)

Vậy tập nghiệm của phương trình là \(S=\left\{1;-2\right\}\)

d) \(\left(x-1\right)^3+\left(2x+3\right)^3=27x^3+8\)

\(\Leftrightarrow x^3-3x^2+3x-1+8x^3+36x^2+54x+27=27x^3+8\)

\(\Leftrightarrow9x^3+33x^2+57x+26=27x^3+8\)

\(\Leftrightarrow18x^3-33x^2-57x-18=0\)

\(\Leftrightarrow18x^3-54x^2+21x^2-63x+6x-18=0\)

\(\Leftrightarrow18x^2\left(x-3\right)+21x\left(x-3\right)+6\left(x-3\right)=0\)

\(\Leftrightarrow\left(x-3\right)\left(18x^2+21x+6\right)=0\)

\(\Leftrightarrow\left(x-3\right)\left(18x^2+9x+12x+6\right)=0\)

\(\Leftrightarrow\left(x-3\right)\left[9x\left(2x+1\right)+6\left(2x+1\right)\right]=0\)

\(\Leftrightarrow\left(x-3\right)\left(2x+1\right)\left(9x+6\right)=0\)

\(\Leftrightarrow\)\(x-3=0\)

hoặc  \(2x+1=0\)

hoặc  \(9x+6=0\)

\(\Leftrightarrow\)\(x=3\)

hoặc  \(x=-\frac{1}{2}\)

hoặc \(x=-\frac{2}{3}\)

Vậy tập nghiệm của phương trình là \(S=\left\{3;-\frac{1}{2};-\frac{2}{3}\right\}\)

\(\left(x+1-\frac{4}{x+1}\right):\frac{x+3}{x^2-2x-3}\)\(=\frac{\left(x+1\right)^2-4}{x+1}.\frac{x^2-2x-3}{x+3}\)

\(=\frac{\left(x+1-2\right)\left(x+1+2\right)}{x+1}.\frac{\left(x+1\right)\left(x-3\right)}{x+3}\)

\(=\frac{\left(x-1\right)\left(x+3\right)}{x+1}.\frac{\left(x+1\right)\left(x-3\right)}{x+3}=\left(x-1\right)\left(x-3\right)=x^2-4x+3\)

\(\left(x+1-\frac{4}{x+1}\right):\frac{x+3}{x^2-2x-3}\)

\(=\frac{\left(x+1\right)^2-4}{x+1}.\frac{\left(x+1\right)\left(x-3\right)}{x+3}\)

\(=\frac{\left(x^2+2x+1-4\right)\left(x-3\right)}{\left(x+3\right)}\)

\(=\frac{\left(x^2+2x-3\right)\left(x-3\right)}{\left(x+3\right)}\)

\(=\frac{\left(x+3\right)\left(x-1\right)\left(x-3\right)}{x+3}\)

\(=\left(x-1\right)\left(x-3\right)\)

\(=x^2-4x+3\)

a) \(bđt\Leftrightarrow a^2+2a< a^2+2a+1\)

\(\Rightarrow0< 1\)(luôn đúng)

b) \(bđt\Leftrightarrow m^2+n^2+2-2m-2n\ge0\)

\(\Leftrightarrow\left(m^2-2m+1\right)+\left(n^2-2n+1\right)\ge0\)

\(\Leftrightarrow\left(m-1\right)^2+\left(n-1\right)^2\ge0\)(đúng)

Dấu "=" khi m = n = 1

c) Áp dụng bđt cô - si với 2 số không âm:

\(a+b\ge2\sqrt{ab}\)

\(\frac{1}{a}+\frac{1}{b}\ge2\sqrt{\frac{1}{ab}}\)

\(\Rightarrow\left(a+b\right)\left(\frac{1}{a}+\frac{1}{b}\right)\ge4\)

Dấu "=" khi a = b

2 hình thức là hình thức gì vậy???

Ta có \(\Leftrightarrow x^4+x^3+x^2+x+1=0\)

\(\Leftrightarrow x^4+x^2+x^3+x+x^2+1=0\)

\(\Leftrightarrow x^2\left(x^2+1\right)x\left(x^2+1\right)+\left(x^2+1\right)=0\)

\(\Leftrightarrow\left(x^2+x+1\right)\left(x^2+1\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x^2+x+1=0\left(ktm\right)\\x^2+1=0\left(ktm\right)\end{cases}}\)

=> Pt vô nghiệm

đpcm.

\(x^4+x^3+x^2+x+1=0\)

\(\Rightarrow\left(x-1\right)\left(x^4+x^3+x^2+x+1\right)=0\)

\(\Rightarrow x^5-1=0\)

\(\Rightarrow x^5=1\)

\(\Rightarrow x=1\)

Nhưng thay vào PT ko đúng nên PT vô nghiệm